Who's familiar with the conversion? - "In 15 years' ship-time they could reach Andromeda

@layman,

I answered the question. You either don't understand the answer, or you are ignoring it.

For any instance in time there is exactly one inertial frame of reference in which you aren't moving. This is what I think you mean by the non-mathematical term "My frame of reference".

So yes, if you are defining the term "my frame of reference" as the frame of reference in which you aren't moving, then yes... in "[your] frame of reference" you are never moving by definition.

What I am patiently trying to explain to you is that the frame of reference in which you aren't moving changes every time you accelerate (in any frame).

That would mean what you are calling "my frame of reference" when start out on earth would be a different frame of reference than the one you call "my frame of reference" when you end up on the spaceship.

Do you understand this? I think is the one of the things you have wrong in your way of thinking about this.

2 Replies

@layman,

Quote:

Let's say I want to calculate it from the way the spaceship "sees" it. Let's say I am now on the ship, instead of the earth, OK. Now that's *my* frame of reference. Under SR, am I moving now?

Just to make this abundantly simple... assuming that by "my frame of reference" you mean the frame of reference in which you are not moving.

The answer is no. You are not moving in either of these two frames of reference (by definition).

1 Reply

@maxdancona,

Quote:

That would mean what you are calling "my frame of reference" when start out on earth would be a different frame of reference than the one you call "my frame of reference" when you end up on the spaceship.

Do you understand this?

Yes, I understand that perfectly. It is presupposed by my question, which you still haven't answered. A simple "yes" or "no" is all that's required. Not an essay full of non sequiturs. The question was:

Quote:

Now that's *my* frame of reference. Under SR, am I moving now?

1. yes? or
2. no?

1 Reply

@maxdancona,

Quote:

The answer is no. You are not moving in either of these two frames of reference (by definition).

OK, thank you. Finally. Now, then...(see next post)

1 Reply

@layman,

Quote:

Just to make this abundantly simple... assuming that by "my frame of reference" you mean the frame of reference in which you are not moving.

The answer is no. You are not moving in either of these two frames of reference (by definition).

I don't know how I can force you to read this. I answered the question very clearly.

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@layman,

The guy on the earth says HE is not moving

The guy on the spaceship says HE is not moving (but rather that the guy on the earth is the one moving).

Which one is right?

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@layman,

Are you just arguing to argue without even bothering to read and at least try to understand?

I already tried to explain this to you.

Max on the last page wrote:

I answered the question. You either didn't understand the answer, or you ignored it.

You can make valid calculations from any frame of reference. They are equally valid. At any instance of time, there is exactly one inertial frame of reference in which you are motionless. It may be more simple to make the calculation from that frame of reference, but any other frame of reference would be valid too.

If we want to talk about mathematics in any more detail than that, we need to be doing math.

1 Reply

@layman,

You can actually experience this without a spaceship. An airplane will suffice.

I can walk in my house with no difficulty. The simplest way to do this is to make all of my calculations in the frame of reference where the house is motionless. I can drop something in my house and it will fall straight down.

A while ago I was in a airplane... basically a metal cylinder with wings which to people on earth was moving forward in a straight line at 380 mile per hour. Yet while I was in the airplane I could assume that the airplane was motionless (in the same way I assumed my house was motionless). While the airplane was cruising at constant speed I could walk around the airplane with no difficulty. I could drop something in the airplane and it would fall straight down.

There is no rule of physics that operates any different in the house frame (assuming my house is motionless) then it does in the airplane frame (assuming the airplane is motionless). The calculations are equally valid.

The math all works with no contradiction, and experiments back up the mathematics.

1 Reply

@maxdancona,

Quote:

I already tried to explain this to you.

You didn't "explain" a damn thing. You did, however, make seemingly incoherent assertions, such as:

Quote:

You can make valid calculations from any frame of reference. They are equally valid

Keep in mind that I'm not asking about math. I'm asking about who, out there is the real world, is moving, if anyone? My question was:

Quote:

Which one is right?

Let me make it easier by giving you multiple choices, eh?

1. The guy on earth is right
2. The guy on the spaceship is right.
3. Both are right
4. Neither is right
5. It cannot be determined who is right.

Which one? If those aren't enough, then state your own answer.

1 Reply

@layman,

Quote:

You can make valid calculations from any frame of reference. They are equally valid. At any instance of time, there is exactly one inertial frame of reference in which you are motionless. It may be more simple to make the calculation from that frame of reference, but any other frame of reference would be valid too.

How many times do I need to keep repeating myself? I don't think you are even trying to understand.

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@maxdancona,

Quote:

The math all works with no contradiction, and experiments back up the mathematics.

I know this, and have repeatedly acknowledged it (at least the math part--but not the "experiment" part). But that's not what I'm asking.

DO YOU UNDERSTAND THAT'S NOT WHAT I'M ASKING? If you did, then you would know it doesn't answer my question.

1 Reply

@layman,

You asked "which frame of reference is valid?".

I answered "they are equally valid". I answered your question.

1 Reply

@maxdancona,

Quote:

I answered "they are equally valid".

For math? Or for people out in the real world?

Which of the following is "equally valid" supposed to assert? 3? 4? 5?

1. The guy on earth is right
2. The guy on the spaceship is right.
3. Both are right
4. Neither is right
5. It cannot be determined who is right

1 Reply

@layman,

I gave an example of how this works in the real world.

Did you read how this works in the airplane example? You have been on a airplane and experienced this yourself, right?

2 Replies

@maxdancona,

Can you answer the question? Which was:

Quote:

Which of the following is "equally valid" supposed to assert? 3? 4? 5?

1 Reply

@layman,

That would be 3. Each is right as calculated in his own frame of reference. The frames of reference are equally valid.

1 Reply

@maxdancona,

Again, when you are in an airplane and I am in my house. To me, the airplane and everything in it is moving at 380 mile per hour in a straight level flight.

You are in the airplane, you would say correctly that the seat you are sitting in is not moving (and every experiment or calculation you could do inside the airplane would agree).

In this case I would be right that the plane (and the seats attached to it) are moving relative to the Earth. And you would be right that the seat you are sitting in in not moving relative to the Plane.

We would both be right.

1 Reply

@maxdancona,

Quote:

I gave an example of how this works in the real world. Did you read how this works in the airplane example?

Yeah, I read it. But it didn't answer my question about the real world. You just gave an answer about math:

Quote:

The math all works with no contradiction

Let me ask this a different way, with example, OK? Assume I want to predict movements of the planets in the solar system. Let's say I look at two optional "frames of reference" for doing this, to wit:

1.The earth. I assume the earth is motionless, and, using a geocentric Ptolemic astronomy, I calculate when the next solar eclipse will occur.

2. The sun. I assume the sun is motionless (and, now, that the earth is moving). Once again I calculate when the next solar eclipse will occur this time using a heliocentric model of the solar system.

Either way, I get the same time and date. Either on is "equally valid" for THAT purpose.

But are they "equally valid" for "real world" purposes? Is just as valid to say that the sun is moving around the earth as it is to say the earth is moving around the sun?

Are those two "equally valid" statements about physical reality?

1 Reply

@layman,

Not exactly. There is a difference between an inertial frame (in which the origin or the reference point is not accelerating) and a non-inertial frame (in which the origin is accelerating). You can use the Earth as an inertial reference frame at any given point of time... but the next moment it will be motionless in a different reference frame.

The Earth and the sun are non-inertial reference frames due to the fact that they are both orbiting a central point. Orbiting a central point is a form of acceleration.

We are about to dive into more advanced mathematics.... but the principle that one person can make a measurement in one reference frame, and another person can make the same measurement in a different reference frames, and they can get different (equally valid) answers.

1 Reply

@maxdancona,

Quote:

There is a difference between an inertial frame (in which the origin or the reference point is not accelerating) and a non-inertial frame (in which the origin is accelerating).

This is a difference, sure. But for purposes of applying the LT and calculations of time dilation, etc., using SR, acceleration is irrelevant. Research "the clock hypothesis," if you think otherwise.

1 Reply

Who's familiar with the conversion? - "In 15 years' ship-time they could reach Andromeda

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